Question 1158074
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    P = {{{C[5]^4*(1/2)^5}}} = {{{5*(1/2)^5}}} = {{{5/32}}}.


Your winning configuration is  (4 heads and 1 tail) of 5 tosses,

and you can get it in 5 different sub-cases. 
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It is a binomial distribution probability problem with &nbsp;n= 5;  &nbsp;k= 4  &nbsp;and &nbsp;&nbsp;p = {{{1/2}}}.